I have a ID field that is 11 chars long 123456789abc
How can I check the length
how can I check position 3 to ensure it's numeric
Select * from table
where ID = position 2 must be 'A-Z'
You can use like with patterns. Something like this:
Select *
from table
where id like '___[0-9]%' and len(id) = #length
Related
I want to know if the 4th integer in the ID, is even, or if its odd.
If the 4th number is even (if the number is either 0,2,4,6,8 I want to put the ID into a new column named 'even'
IF the 4th number is odd, the column should have the name 'Odd'
select ID as 'Female'
from Users2
where ID LIKE '%[02468]'
This shows if any of the numbers are even. I want to specify the 4th number
Try this:
select *, OddOrEven = iif(substring(ID,4,1) in ('0','2','4','6','8') , 'Even', 'Odd') from Users2
This will tell you whether the 4th character is Odd or Even.
This is of course assuming that the 4th character of ID column will be numeric.
To make it permanently part of the table, you can add a computed column as shown below.
alter table Users2
add OddOrEven as iif(substring(ID,4,1) in ('0','2','4','6','8'), 'Even', 'Odd')
Substring the character you are interested in
Convert to an int
Check whether modulus 2 returns 0 (i.e. even).
select id
, case when convert(int,substring(id, 4, 1)) % 2 = 0 then 'Even' else 'Odd' end
from Users;
Example:
select id
, case when convert(int,substring(id, 4, 1)) % 2 = 0 then 'Even' else 'Odd' end
from (values ('4545-4400'), ('4546-4400')) X (id);
Returns
id
4545-4400
Odd
4546-4400
Even
Thats assuming there is always a 4th character. If not you would need to check for it.
You were close, but only need to check a single character against a set of characters:
where Substring( Id, 4, 1 ) like '[02468]'
Note that there is no wildcard (%) in the pattern.
It can be used in an expression like:
case when Substring( Id, 4, 1 ) like '[02468]' then 'Even' else 'Odd' end as Oddity
How can I select records where in the column Value the 5th character is letter A?
For example the following records:
ID Value
-------------------------
1 1234A5636A6363
2 1234A4343B6363
3 1234B5353A6363
if I run
select * from table
where Value like '%A%'
this will return all records
but all I want is the first 2 where the 5th character is A, regardless if there are more A characters in the text or not
select *
from your_table
where substring(Value, 5, 1) = 'A'
The LIKE operator, in addition to %, which matches any number of any character, can use _, which matches any one single character. You may try:
SELECT *
FROM yourTable
WHERE Value LIKE '____A%'; -- 4 underscores here
use like below by using _(underscore)
LIKE '____A%'
SQL Server
select *
from YourTableName
where CHARINDEX('A', ColumnName) = 5
Note:- This finds where string 'A' starts at position 5
AND specify Your ColumnName
Is there a way to use Pattern Matching with SQL LIKE, to match a variable number of characters with an upper limit?
For example, I have one column which can have "correct values" of 2-10 numbers, anything more than 10 and less than 2 is incorrect.
If you really want to use like you can do:
where col like '__' or col_like '___' or . . .
or:
where col like '__%' and -- at least two characters
col not like '_________%' -- not 9 characters
The more typical method would be:
where len(col) between 2 and 10
If you want to ensure they are numbers:
where len(col) between 2 and 10 and
col not like '%[^0-9]%'
You could make a function to do this the long way of inspecting each character like below:
DECLARE #TempField varchar(max)= '1v3Abc567'
DECLARE #FieldLength int = LEN(#TempField)
DECLARE #CountNumeric int = 0
WHILE #FieldLength > 0
BEGIN
SELECT #CountNumeric = #CountNumeric + ISNUMERIC( SUBSTRING (#TempField,#FieldLength,1))
SET #FieldLength = #FieldLength - 1
END
SELECT #CountNumeric NumericCount
You can use Regex.
SELECT * FROM mytable WHERE mycol LIKE '%[0-9]{2,10}%';
should do it (that's off the top of my head, so double-check!
I need to create a SQL Query.
This query need to select from a table where a column contains regular expression.
For example, I have those values:
TABLE test (name)
XHRTCNW
DHRTRRR
XHRTCOP
CPHCTPC
CDDHRTF
PEOFOFD
I want to select all the data who have "HRT" after 1 char (value 1, 2 and 3 - Values who looks like "-HRT---") but not those who might have "HRT" after 1 char (value 5).
So I'm not sure how to do it because a simple
SELECT *
FROM test
WHERE name LIKE "%HRT%"
will return value 1, 2, 3 and 5.
Sorry if I'm not really clear with what I want/need.
You can also change the pattern. Instead of using % which means zero-or-more anything, you can use _ which means exactly one.
SELECT * FROM test WHERE name like '_HRT%';
You can use substring.
SELECT * FROM test WHERE substring(name from 2 for 3) = 'HRT'
Are the names always 7 letters? Do:
SELECT substring (2, 4, field) from sometable
That will just select the 2-4th characters and then you can use like "%HRT"
How can I write a select statement to select only integers (and nothing more) from a char column in SQL Server. For example, my table name is POWDER with 2 columns, ID (int) and Name(char (5))
ID Name
-- ----------
1 AXF22
2 HYWWW
3 24680
4 8YUH8
5 96635
I want to be able to select only those rows that contain an integer and nothing more (ID 3 and ID 5 in this example)
If I try:
SELECT *
FROM POWDER
WHERE Name LIKE '[0-9]%'
...it will return:
ID Name
-- ----------
3 24680
4 8YUH8
5 96635
Any ideas how to get the rows containing just integers?
SELECT * FROM POWDER WHERE IsNumeric(Name) = 1
IsNumeric returns 1 for some other characters that are valid in numbers, such as + and - and $ but for your input you should be fine.
Try this:
SELECT * FROM Table WHERE Name LIKE '[0-9]%%'
To avoid issues with ISNUMERIC and all spaces, -, +, . etc, use the fact that the column is char(5)
SELECT *
FROM POWDER
WHERE Name LIKE '[0-9][0-9][0-9][0-9][0-9]'
Edit: for any number of characters. Double negative...
SELECT *
FROM POWDER
WHERE Name NOT LIKE '%[^0-9]%'
Use positive and negative checks to make sure we have an integer: It must contain a digit. Only digits and spaces are allowed. No spaces are allowed between digits.
SELECT *
FROM POWDER
WHERE Name LIKE '%[0-9]%'
AND Name NOT LIKE '%[^0-9 ]%'
AND Name NOT LIKE '%[0-9]% %[0-9]%'
Try:
SELECT *
FROM POWDER
WHERE Name patindex ('%[a-z]%',name) != 0
The last one is the best,kind of works really really well.
SELECT * FROM POWDER
WHERE Name LIKE '%[0-9]%'
AND Name NOT LIKE '%[^0-9 ]%'
AND Name NOT LIKE '%[0-9]% %[0-9]%'